leetcode之Peeking Iterator_python peek iter-优快云博客

本文链接：https://blog.youkuaiyun.com/howtogetout/article/details/49591579

本文介绍了一种在迭代器中实现peek功能的方法，通过预加载所有元素到列表中，并使用索引来模拟peek操作，同时保持next操作的正常进行。

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本题是可以peek一下下一个数字，但是不改变next的状态。所以保存一个中间变量，当需要peek的时候就是用那个变量，而需要next的时候则中间变量+ 1，这样就能获得本身需要的功能了。代码如下：

# Below is the interface for Iterator, which is already defined for you.
#
# class Iterator(object):
#     def __init__(self, nums):
#         """
#         Initializes an iterator object to the beginning of a list.
#         :type nums: List[int]
#         """
#
#     def hasNext(self):
#         """
#         Returns true if the iteration has more elements.
#         :rtype: bool
#         """
#
#     def next(self):
#         """
#         Returns the next element in the iteration.
#         :rtype: int
#         """

class PeekingIterator(object):
    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.item = iterator
        self.items = []
        while self.item.hasNext():
            self.items.append(self.item.next())
        print self.items
        self.num = 0

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        if self.num >= len(self.items):
            pass
        else:
            return self.items[self.num]
        

    def next(self):
        """
        :rtype: int
        """
        if self.num >= len(self.items):
            pass
        else:
            self.num = self.num + 1
            return self.items[self.num - 1]
        

    def hasNext(self):
        """
        :rtype: bool
        """
        if self.num >= len(self.items):
            return False
        else:
            return True